3 equations with 4 unknowns and information about the unknowns' types My dad just sent me a math problem on email and asked if I could solve it, but frankly, I have no idea where to start. I've got 3 equations with 4 unknowns and some information about the unknowns.
First, the equations:
$$2(z-1)-x=55$$
$$4xy-8z=12$$
$$a(y+z)=11$$
I am supposed to determine the 2 greatest real number values of $a$, where $x,y,z$ are positive natural numbers.
The problem is driving me nuts, so any help will be greatly appreciated (even just a hint). Thanks in advance.
 A: You have from the two first equations :
$ 2z-x = 57$ and $xy-2z = 3$ 
thus $$x(y-1) = 60 = 2^2\times3\times5$$
Therefore since $x$ divides 60,  $$x\in\{1,2,3,4,5,6,10,15,18,20,30,60\}$$ and thus $$xy=60+x \in \{61,62, 63, 64, 65, 66, 70, 75, 78, 80, 90, 120\}$$
However from $xy - 2z = 3$ we see that $xy$ is odd, thus $xy \in\{61, 63, 65, 75\}$ and thus $x\in\{1, 3, 5, 15\}$ ie $y\in\{61, 21, 13, 5\}$
and $z = {(xy-3)\over2} \in \{29, 30, 31, 36 \}$
Therefore the only solutions to the first equations are  $$(x,y,z)\in\{(1, 61,29 ),(3, 21, 30), (5, 13, 31), (15, 5, 36)\}$$
Then $$a = {11\over y+z} \in \{{11\over 90}, {11\over 51}, {11\over 44}, {11\over 41}\}$$
I let you figure out which one is the greatest
A: You can solve the first equation for $x$ and plug into the second.
$$2(z-1)-x=55\\
x=2(z-1)-55\\
8zy-8y-8z-220y=12\\
2z(y-1)-57y=3\\
z=\frac {3+57y}{2(y-1)}=\frac{30}{y-1}+\frac {57}2$$
To make $z$ integral, we must have $y-1$ a multiple of $4$ but not $8$ and divisible into $60$, so $y$ can equal $5,13,21,61$ giving the solutions 
$$ \begin {array} {c c c c}x&y&z&a\\15&5&36&\frac {11}{41}\\
5&13&31&\frac {11}{44}\\
3&21&30&\frac {11}{51}\\1&61&29&\frac {11}{90}\end {array}$$
A: The thing that jumps out at me is
$4xy - 8z = 12$ means
$xy -2z = 3$  .... oh, for a second there, I thought that bounded things.
So well...
it gives a relation between $z$ and $xy$ as $z = \frac{xy -3}2$ which means $xy-3$ is even which means $xy$ is odd so $x$ and $y$ are odd.
We also have $2(z-1)-x=55$ so $2(\frac{xy-3}2 - 1) - x = 55$ so $xy-3 -2 - x = 55$
so $xy - x = 60$ so $x(y-1) = 60 = 4*3*5$ but both $x,y$ are odd $y-1$ is even. 
So we have:
$(x,y-1) = \{(15,4)(5,12)(3,20)(1,60)\}$
And $(x,y,z=\frac{xy-3}2) = \{(15,5,36),(5,13,31),(3,21,30),(1,61,29)\}$
So $y + z = \{41, 44, 51, 90\}$
So $a(y+z) = 11 \implies a = \frac{11}{y+z} = \{\frac{11}{41},\frac{11}{44},\frac{11}{51},\frac{11}{90}\}$
so the to greatest are $a = 11/41$ and $x,y,z =15,5,36$ and $a= 1/4$ and $x,y,z = 5,13,31$.
Check: 
$2(z−1)−x=55; 2(36 - 1) - 15 = 70-15=55;2(31-1)-5=60-5=55$
$4xy−8z=12;4*15*5 -8*36 = 300 - 288 =12; 4*5*13-8*31= 260 - 248 = 12$
$a(y+z)=11;\frac{11}{41}*(5+36) = 11; \frac 14*(13+31) = \frac{44}4 = 11$. 
